本文主要研究如下含多个奇性项和临界指数增长的非线性次椭圆型方程的非平凡解在奇点处的渐近性质

$ \begin{equation} \left\{ \begin{aligned} &-(\Delta _x + |x|^{2\alpha }\Delta _y) u - \mu \frac{\prod_{i = 1}^{k}\psi_{i}^{2}u}{\prod_{i = 1}^{k} d(z, a_{i})^2} = f(z, u), \quad &z\in\Omega , \\ &\quad u = 0, &z\in\partial \Omega, \end{aligned} \right. \end{equation} $

(1.1)

其中$ \Omega $是$ \mathbb {R}^N: = \mathbb {R}^m\times \mathbb {R}^n $中的开集, $ k\geq 2 $, $ a_{i}\in \Omega $ ($ i = 1, 2, \cdots, k $)且互不相同, $ 0\leq \mu<\mu_{G}: = (\frac{Q-2}{2})^{2} $, 这里$ (\frac{Q-2}{2})^{2} $是最佳Hardy常数; $ \alpha>0 $, $ \Delta _x + |x|^{2\alpha }\Delta _y $是Grushin型算子, $ Q = m+(\alpha+1)n $是空间$ \mathbb {R}^m\times \mathbb {R}^n $的齐次维数.对任意的$ z = (x, y) $, $ a_{i} = (a_{i}^{1}, a_{i}^{2})\in \Omega\subset \mathbb{R}^{m}\times\mathbb{R}^{n} $, $ \psi_{i} = |\nabla_{\alpha}d(z, a_{i})| = (\frac{|x-a_{i}^{1}|}{d(z, a_{i})})^{\alpha} $, 其中$ d(z, a_{i}) $表示在Grushin向量场意义下点$ z $与点$ a_{i} $之间的距离, 其定义如下:

$ \begin{equation} d(z, a_{i}) = \Big(|x-a_{i}^{1}|^{2(\alpha +1)}+(\alpha +1)^2|y-a_{i}^{2}|^2\Big)^{\frac{1}{2(\alpha +1)}}. \end{equation} $

(1.2)

此外, 非线性项$ f: \Omega \times \mathbb {R}\to\mathbb {R} $是Carathéodory函数, 且满足

$ \begin{equation} |f(z, t)|\leq C(|t|^{q-1}+ |t|^{2^{*}-1}), \quad \forall (z, t)\in \Omega\times\mathbb{R}, \end{equation} $

(1.3)

其中$ C>0 $是常数, $ 2^* = \frac{2Q}{Q-2} $是临界Sobolev指数.

记

$ X_i = \frac{\partial }{\partial x_i} \, \, ( i = 1, \ldots , m), \quad X_{m+j} = |x|^\alpha \frac{\partial }{\partial y_j}\, \, (j = 1, \ldots , n). $

则Grushin梯度为$ \nabla _\alpha = (X_1, \ldots , X_N) $, 由此Grushin型算子可表示为

$ \begin{equation} \Delta _x + |x|^{2\alpha }\Delta _y = \text{div}_{\alpha}(\nabla_{\alpha}) = \sum\limits_{i = 1}^{m}X_{i}^{2}+\sum\limits_{j = 1}^{n}X_{m+j}^{2}. \end{equation} $

(1.4)

显然:当$ x\neq 0 $时, 该算子是椭圆型的, 并且在流形$ \{0\}\times \mathbb {R}^n $上是退化的; 当$ \alpha $是非负整数时, Grushin算子是Hörmander型的.其它相关知识可以参见[1-5]等.

对任意的$ \Omega \subset \mathbb {R}^N $, 范数$ \| u\|_{D^{1, 2}_{\alpha}} = (\int _{\Omega } |\nabla _\alpha u|^2 dz)^{\frac{1}{2}} $下的$ C^\infty _0(\Omega ) $完备化空间记为$ D_{\alpha}^{1, 2}(\Omega) $.对$ \forall u\in D_{\alpha}^{1, 2}(\Omega) $, 设方程(1.1)的能量泛函为

$ \begin{equation} I(u) = \frac{1}{2}\int _{\Omega } |\nabla _\alpha u|^2 dz- \frac{\mu}{2}\int _{\Omega } \frac{\prod_{i = 1}^{k}\psi_{i}^{2}|u|^{2}}{\prod_{i = 1}^{k} d(z, a_{i})^2}dz- \int_{\Omega}F(z, u)dz, \end{equation} $

(1.5)

其中$ F(z, t) = \int_{0}^{t}f(z, \eta)d\eta $.若$ u\in D_{\alpha}^{1, 2}(\Omega) $是方程(1.1)的解, 则对任意$ v\in C^{\infty}_{0}(\Omega) $有

$ \begin{equation} \langle I'(u), v\rangle = \int _{\Omega } \nabla _\alpha u \nabla _\alpha v dz- \mu\int _{\Omega } \frac{\prod_{i = 1}^{k}\psi_{i}^{2}uv}{\prod_{i = 1}^{k} d(z, a_{i})^2}dz- \int_{\Omega}f(z, u)vdz. \end{equation} $

(1.6)

为了应用变分方法处理(1.1), 我们需要如下的Hardy不等式:

$ \begin{equation} \int _{\Omega} |\nabla _\alpha u|^2 dz \geq \mu_{G} \int _{\Omega } \psi ^2 \frac{|u|^2}{d(z, a)^2}dz, \quad \forall u\in C_0^{\infty }(\varOmega ), \end{equation} $

(1.7)

其中$ \mu_{G } = ( \frac{Q-2}{2} )^2 $是Grushin向量场下Hardy型不等式的最佳常数, $ a\in\Omega $.

在文[6]中作者研究了如下含多个奇性项的退化次椭圆问题:

$ \begin{equation} \left\{ \begin{aligned} &-\Delta_{G}u -\sum\limits_{i = 1}^{k}\mu_{i} \frac{\psi ^2}{ d(z, a_{i})^2} u = f(z, u), \qquad &\quad z\in\Omega , \\ &u = 0, & z\in\partial \Omega, \end{aligned} \right. \end{equation} $

(1.8)

其中$ \Omega $是Carnot群上的开集, $ -\Delta_{G} $是Carnot群上的Laplace算子, $ k\geq 2 $, 实数$ a_{i}\in \Omega $ ($ i = 1, 2, \cdots, k $)互不相同, $ \mu_{i}\geq 0 $且满足$ \sum\limits_{i = 1}^{k}\mu_{i}<\mu_{C} = (\frac{C-2}{2})^{2} $, $ C $是Carnot群的齐次维数.方程(1.8)与如下Carnot群上的Hardy型不等式有关:

$ \begin{equation} \int _{\Omega} |\nabla _{G} u|^2 dz \geq\sum\limits_{i = 1}^{k} \mu_{i} \int _{\Omega } \psi ^2 \frac{|u|^2}{d(z, a_{i})^2}dz, \quad \forall u\in S_0^{1}(\Omega ). \end{equation} $

(1.9)

类似于Carnot群上的证明方法, 可以建立Grushin向量场下的多奇性项Hardy型不等式:

$ \begin{equation} \int _{\Omega} |\nabla _{\alpha} u|^2 dz \geq\sum\limits_{i = 1}^{k} \mu_{i} \int _{\Omega } \psi_{i}^2 \frac{|u|^2}{d(z, a_{i})^2}dz, \quad \forall u\in D_{\alpha}^{1, 2}(\Omega ). \end{equation} $

(1.10)

结合(1.7), (1.10)式, 可以得到如下的广义Hardy型不等式:

$ \begin{equation} \int _{\Omega} |\nabla _\alpha u|^2 dz \geq \mu^{*}_{G} \int _{\Omega } \frac{\prod_{i = 1}^{k} \psi_{i}^2|u|^2}{ \prod_{i = 1}^{k}d(z, a_{i})^2}dz, \quad \forall u\in D_{\alpha}^{1, 2}(\Omega ), \end{equation} $

(1.11)

其中$ \mu^{*}_{G} $是广义Hardy型不等式的最佳常数.不等式(1.11)的证明将在第二节中给出.

假设$ k\geq 2 $, 对任意的$ i = 1, 2, \cdots, k $, 定义

$ \mu_{i} = \frac{\mu}{\prod\limits_{j = 1, j\neq i}^{k}d(a_{i}, a_{j})^{2}}. $

利用Moser迭代技巧, 本文讨论了方程(1.1)的非平凡解在奇点处的渐近性质.结论如下.

定理1.1   设$ \Omega $是$ \mathbb {R}^m\times\mathbb {R}^n $中的开集, $ a_{j}\in \Omega $ $ (j = 1, 2, \cdots, k) $且互不相等. $ 0<\mu<\mu^{*}_{G} $, $ \sum\limits_{i = 1}^{k}\mu_{i}<(\frac{Q-2}{2})^{2} $.若$ u\in D_{\alpha }^{1, 2}(\Omega ) $是方程(1.1)的非平凡弱解, 则

(i) 存在常数$ C $, $ \rho_{i}>0 $使得

$ \begin{equation} |u(z)|\le \frac{C}{d(z, a_{i})^{\frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\mu_{i} }}}, \quad \forall z\in B_{d}(a_{i}, \rho_{i})\backslash \{a_{i}\}, \end{equation} $

(1.12)

其中$ B_{d}(a_{i}, \rho_{i}) $表示在距离$ d $的意义下以$ a_{i} $为圆心, 以$ \rho_{i} $为半径的球, 且$ a_{j}\not\in B_{d}(a_{i}, \rho_{i}) $, $ i, j = 1, 2, \cdots , k $, $ i\neq j $.

(ii) 存在常数$ C>0 $使得

$ \begin{equation} |u(z)|\le C\sum\limits_{i = 1}^{k}\frac{1}{d(z, a_{i})^{\frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\mu_{i} }}}, \quad \forall z\in \Omega\backslash \{a_{1}, a_{2}, \cdots a_{k}\}. \end{equation} $

(1.13)

在欧式空间中, 关于Laplace算子问题的相关结论可参考[7].利用Moser迭代和分析技巧, 我们在第二节给出定理1.1的证明.对于退化的Grushin型算子和Carnot群上的次椭圆算子而言, 该结论依然是新的.

首先给出广义Hardy型不等式(1.11)的证明.

不等式(1.11)的证明 令$ \rho = \min\{\frac{1}{2}d(a_{i}, a_{l}), d(a_{i}, \partial\Omega)\} $, $ i, \, l = 1, 2, \cdots, k $, $ i\neq l $.则$ B_{d}(a_{i}, \rho)\cap B_{d}(a_{l}, \rho) = \emptyset $, 其中$ B_{d}(a, \rho) = \{x:x\in \Omega, d(x, a)<\rho\} $.由不等式(1.7)和$ \psi_{i}<1 $ ($ i = 1, 2, \cdots, k $)可得

$ \begin{equation} \begin{aligned} \int _{\Omega } \frac{\prod_{i = 1}^{k}\psi_{i}^2|u|^2}{\prod_{i = 1}^{k}d(z, a_{i})^2}dz & = \sum\limits_{i = 1}^{k}\int _{B_{d}(a_{i}, \rho) } \frac{\prod_{i = 1}^{k}\psi_{i}^2|u|^2}{\prod_{i = 1}^{k}d(z, a_{i})^2}dz+ \int _{\Omega\backslash \bigcup\limits_{i = 1}^{k}B_{d}(a_{i}, \rho) } \frac{\prod_{i = 1}^{k}\psi_{i}^2|u|^2}{\prod_{i = 1}^{k}d(z, a_{i})^2}dz\\ &\leq \frac{1}{\rho^{2(k-1)}} \sum\limits_{i = 1}^{k}\int _{B_{d}(a_{i}, \rho) } \psi_{i}^2 \frac{|u|^2}{d(z, a_{i})^2}dz+ \frac{1}{\rho^{2k}} \int _{\Omega\backslash \bigcup\limits_{i = 1}^{k}B_{d}(a_{i}, \rho) } |u|^2dz\\ &\leq \frac{1}{\rho^{2(k-1)}}\sum\limits_{j = 1}^{k}\int _{\Omega} \frac{|u|^2}{d(z, a_{j})^2}dz+\frac{1}{\rho^{2k}}\int _{\Omega} |u|^{2}dz\\ &\leq C(\Omega, k, \rho)\|u\|_{D^{1, 2}_{\alpha}}^{2}. \end{aligned} \end{equation} $

(2.1)

因此, 不等式(1.11)得证.上述结论表明广义Hardy型不等式的最佳常数可如下定义:

$ \mu^{*}_{G} = \inf\limits_{u\in D^{1, 2}_{\alpha}(\Omega)\backslash\{0\}} \frac{\|u\|_{D^{1, 2}_{\alpha}}^{2}}{\int _{\Omega } \frac{ \prod_{i = 1}^{k}\psi_{i}^2 |u|^2}{\prod_{i = 1}^{k}d(z, a_{i})^2}dz}\geq \frac{1}{C(\Omega, k, \rho)}>0. $

为了估计方程(1.1)的非平凡解在原点处的渐近性质, 我们需要如下的$ L^p $估计.

引理2.2   设$ \Omega \subset \mathbb{R}^m\times \mathbb{R}^{n} $是$ 0 $的有界领域, $ 0\leq \mu <\mu^{*}_{G} $, $ \sum\limits_{i = 1}^{k}\mu_{i}<\mu_{G} $.假设$ V\in L^{\frac{Q}{2}}(\Omega ) $, $ g\in L^q(\Omega ) $, $ q\geq 2 $, 常数$ \nu $使得线性算子$ -(\Delta_{x}+|x|^{2\alpha}\Delta_{y}) - \mu \frac{\prod_{i = 1}^{k}\psi_{i}^2}{\prod_{i = 1}^{k}d(z, a_{i})^2}-V +\nu $是正定的.若$ u\in D_{\alpha }^{1, 2}(\Omega ) $是方程

$ \begin{equation} -(\Delta_{x}+|x|^{2\alpha}\Delta_{y}) u - \mu \frac{\prod_{i = 1}^{k}\psi_{i}^2}{\prod_{i = 1}^{k}d(z, a_{i})^2}u-V u +\nu u = g, \quad z\in \Omega, \end{equation} $

(2.2)

的非平凡解, 则$ u\in L^p(\Omega ) $, $ \forall 1\leq p\leq p_{\text{lim}} = 2^* \min \left\{ \frac{q}{2}, \frac{\frac{Q-2}{2}}{\frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\sum\limits_{i = 1}^{k}\mu_{i} }}\right\} . $

证明  引理2.2的证明与[1, 命题3.2]或[7, 引理3.1]证明过程类似.此处略.

为了研究方程解在奇点处的渐近性质, 我们将$ \Omega $做如下分解

$ \Omega = \Omega_{1}\cup \Omega_{2}\cup \cdots \cup \Omega_{k}, $

并且$ a_{i}\in \Omega_{i} $, $ a_{j}\not\in\overline{\Omega}_{i} $ $ (i\neq j) $, $ i = 1, 2, \cdots, k $.

定理1.1的证明  设$ u\in D_{\alpha }^{1, 2}(\Omega ) $是方程(1.1)的解.令

$ \begin{equation} v: = d(z, a_{i})^{\beta }u, \end{equation} $

(2.3)

其中$ \beta = \frac{Q-2}{2}-\sqrt{(\frac{Q-2}{2})^{2}-\mu_{i}} $是方程$ \beta^{2}-\beta(Q-2)+\mu_{i} = 0 $的解.则由Hardy不等式可知$ v\in D_{\alpha}^{1, 2}(\Omega , d(z, a_{i})^{-2\beta}d z) $, 并且

$ \begin{equation} \begin{aligned} (\Delta_{x}+|x|^{2\alpha}\Delta_{y})u & = (\Delta_{x}+|x|^{2\alpha}\Delta_{y})(d(z, a_{i})^{-\beta}v)\\ & = v[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})d(z, a_{i})^{-\beta}] +d(z, a_{i})^{-\beta}[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})v]\\ &\quad +2\langle \nabla_{\alpha} d(z, a_{i})^{-\beta}, \nabla_{\alpha} v\rangle, \end{aligned} \end{equation} $

(2.4)

其中

$ \begin{equation} \begin{aligned} (\Delta_{x}+|x|^{2\alpha}\Delta_{y})d(z, a_{i})^{-\beta} & = \psi_{i}^{2}[\beta(\beta+1)d(z, a_{i})^{-\beta-2}+\frac{Q-1}{d(z, a_{i})}(-\beta)d(z, a_{i})^{-\beta-1}]\\ & = \psi_{i}^{2}[\beta(\beta+1)d(z, a_{i})^{-\beta-2}-\beta(Q-1)d(z, a_{i})^{-\beta-2}]\\ & = \psi_{i}^{2}[\beta^{2}-\beta(Q-2)]d(z, a_{i})^{-\beta-2}.\\ \end{aligned} \end{equation} $

(2.5)

从而将$ u = d(z, a_{i})^{-\beta}v $带入方程(1.1)的左边, 由(2.4), (2.5)式可得

$ \begin{equation} \begin{aligned} &-(\Delta_{x}+|x|^{2\alpha}\Delta_{y})u- \mu\frac{\prod_{i = 1}^{k}\psi_{i}^2 u}{\prod_{i = 1}^{k}d(z, a_{i})^{2}}\\ = &-v\psi_{i}^{2}[\beta^{2}-\beta(Q-2)]d(z, a_{i})^{-\beta-2} -d(z, a_{i})^{-\beta}[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})v]\\ & -2\langle \nabla_{\alpha} d(z, a_{i})^{-\beta}, \nabla_{\alpha} v\rangle -\mu\frac{\prod_{i = 1}^{k}\psi_{i}^2}{\prod_{i = 1}^{k}d(z, a_{i})^{2}}d(z, a_{i})^{-\beta}v\\ = &-v\psi_{i}^{2}[\beta^{2}-\beta(Q-2)+\mu_{i}]d(z, a_{i})^{-\beta-2} +\frac{\mu_{i}\psi_{i}^{2}v}{d(z, a_{i})^{\beta+2}} -\mu\frac{\prod_{i = 1}^{k}\psi_{i}^2}{\prod_{i = 1}^{k}d(z, a_{i})^{2}}d(z, a_{i})^{-\beta}v\\ & -d(z, a_{i})^{-\beta}[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})v] -2\langle \nabla_{\alpha} d(z, a_{i})^{-\beta}, \nabla_{\alpha} v\rangle\\ = &-d(z, a_{i})^{-\beta}[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})v] -2\langle \nabla_{\alpha} d(z, a_{i})^{-\beta}, \nabla_{\alpha} v\rangle\\ &+\frac{1}{d(z, a_{i})^{\beta}} \Big(\frac{\mu_{i}\psi_{i}^{2}}{d(z, a_{i})^{2}} -\frac{\mu \prod_{i = 1}^{k}\psi_{i}^2}{\prod_{i = 1}^{k}d(z, a_{i})^{2}}\Big)v. \end{aligned} \end{equation} $

(2.6)

结合$ f(z, u) = f(z, d(z, a_{i})^{-\beta}v) $及(1.1), (2.6)可得

$ \begin{equation} \begin{aligned} &-d(z, a_{i})^{-\beta}[(\Delta_{x}+|x|^{2\alpha}\Delta_{y})v] -2\langle \nabla_{\alpha} d(z, a_{i})^{-\beta}, \nabla_{\alpha} v\rangle +\frac{1}{d(z, a_{i})^{\beta}} \Big(\frac{\mu_{i}\psi_{i}^{2}}{d(z, a_{i})^{2}} -\frac{\mu \prod_{i = 1}^{k}\psi_{i}^2}{\prod_{i = 1}^{k}d(z, a_{i})^{2}}\Big)v\\ = &f(z, d(z, a_{i})^{-\beta}v). \end{aligned} \end{equation} $

(2.7)

在(2.7)式两边同时乘上$ d(z, a_{i})^{-\beta} $, 可得

$ \begin{equation} \begin{aligned} -\text{div}_{\alpha } \left( d(z, a_{i})^{-2\beta }\nabla _\alpha v \right) = &d(z, a_{i})^{-\beta } f(z, d(z, a_{i})^{-\beta } v )\\& -\frac{1}{d(z, a_{i})^{2\beta}} \Big(\frac{\mu_{i}\psi_{i}^{2}}{d(z, a_{i})^{2}} -\frac{\mu \prod_{i = 1}^{k}\psi_{i}^2}{\prod_{i = 1}^{k}d(z, a_{i})^{2}}\Big)v, \quad z \in \Omega\backslash \{a_{1}, a_{2}, \cdots, a_{k}\}. \end{aligned} \end{equation} $

(2.8)

取$ \rho_{i}>0 $, 使得$ B_{d}(a_{i}, \rho_{i})\subset\Omega $, 并且$ a_{j}\not\in\overline{B}_{d}(a_{i}, \rho_{i}) $, $ i, \, j = 1, 2, \cdots, k $, $ i\neq j $.令$ \rho\in (0, \rho_{i}) $.定义截断函数$ \eta_{i} \in C_{0}^{\infty}( B_{d}(a_{i}, \rho), [0, 1]) $, 使其对$ \forall z\in B_{d}(a_{i}, r) $有$ \eta(z) = 1 $, $ |\nabla_{\alpha}\eta|\leq \frac{4}{\rho-r} $, 其中$ r\in (0, \rho) $.定义函数

$ \begin{equation} v_{L} = \min {\{|v|\, , \, L\}}, \qquad\quad \phi: = \eta^2 v v_{L}^{2(s-1)}, \end{equation} $

(2.9)

其中$ L $, $ s>1 $在后面给定.将试验函数$ \phi $代入(2.8)式得

$ \begin{equation} \begin{aligned} \int_{\Omega}\langle\nabla _\alpha v, \nabla_{\alpha}\phi\rangle d(z, a_{i})^{-2\beta } dz & = \int_{\Omega} d(z, a_{i})^{-\beta } f(z, d(z, a_{i})^{-\beta } v )\phi dz\\ &\quad-\int_{\Omega} \frac{v\phi}{d(z, a_{i})^{2\beta}} \Big(\frac{\mu_{i}\psi_{i}^{2}}{d(z, a_{i})^{2}} -\frac{\mu \prod_{i = 1}^{k}\psi_{i}^2}{\prod_{i = 1}^{k}d(z, a_{i})^{2}}\Big)dz. \end{aligned} \end{equation} $

(2.10)

将$ \phi $的表达式代入(2.10)式左边, 计算可得

$ \begin{equation} \begin{aligned} \int _{\Omega} \frac{\langle\nabla _\alpha v, \nabla_{\alpha}\phi\rangle}{d(z, a_{i})^{2\beta }} dz = &\int _{\Omega}\frac{2\eta v v_{L}^{2( s-1) } \langle \nabla _{\alpha}\eta , \nabla _{\alpha} v \rangle}{d(z, a_{i})^{2\beta}} dz\\ &+\int _{\Omega}\frac{\eta ^2v_{L}^{2( s-1) }| \nabla _{\alpha} v| ^2}{d(z, a_{i})^{2\beta}} dz +2( s-1) \int _{\Omega}\frac{\eta^2 v_{L}^{2( s-1) } | \nabla _{\alpha} v_{L}| ^2}{d(z, a_{i})^{2\beta}} dz. \end{aligned} \end{equation} $

(2.11)

由Young不等式, 对充分小的$ \varepsilon>0 $, 存在$ C_{1}(\varepsilon)>0 $使得

$ \begin{equation} \begin{aligned} \Big|\int _{\Omega} \frac{2\eta v v_{L}^{2( s-1) } \langle \nabla _{\alpha}\eta , \nabla _{\alpha} v \rangle}{d(z, a_{i})^{2\beta}} dz\Big| &\leq \varepsilon \int _{\Omega} \frac{\eta^{2} v_{L}^{2( s-1) } |\nabla_{\alpha}v|^{2}}{d(z, a_{i})^{2\beta}} dz+C_{1}(\varepsilon)\int _{\Omega} \frac{ v^{2} v_{L}^{2( s-1) } |\nabla_{\alpha}\eta|^{2}}{d(z, a_{i})^{2\beta}} dz \end{aligned} \end{equation} $

(2.12)

从而

$ \begin{equation} \begin{aligned} \int _{\Omega} \frac{\langle\nabla _\alpha v, \nabla_{\alpha}\phi\rangle}{d(z, a_{i})^{2\beta }} dz \geq& (1-\varepsilon)\int _{\Omega}\frac{\eta ^2v_{L}^{2( s-1) }| \nabla _{\alpha} v| ^2}{d(z, a_{i})^{2\beta}} dz +2( s-1) \int _{\Omega}\frac{\eta^2 v_{L}^{2( s-1) } | \nabla _{\alpha} v_{L}| ^2}{d(z, a_{i})^{2\beta}} dz\\ &-C_{1}(\varepsilon)\int _{\Omega} \frac{ v^{2} v_{L}^{2( s-1) } |\nabla_{\alpha}\eta|^{2}}{d(z, a_{i})^{2\beta}} dz. \end{aligned} \end{equation} $

(2.13)

将$ \phi $的表达式代入(2.10)式右边第一项中, 利用函数$ f $满足的条件(1.3)可得

$ \begin{equation} \begin{aligned} \int _{\Omega }\frac{f(z, d(z, a_{i})^{-\beta } v) \phi}{d(z, a_{i})^{\beta }} dz & = \int _{\Omega }\frac{f(z, d(z, a_{i})^{-\beta } v) \eta^2vv_{L}^{2(s-1)}}{d(z, a_{i})^{\beta }} dz\\ &\leq \int _{\Omega }\frac{(C(|d(z, a_{i})^{-\beta } v|^{q-1}+|d(z, a_{i})^{-\beta } v|^{2^*-1})) \eta^2vv_{L}^{2(s-1)} }{d(z, a_{i})^{\beta }}dz\\ & = C\left(\int _{\Omega }\frac{|v|^{q}\eta^2 v_{L}^{2(s-1)}}{d(z, a_{i})^{q\beta }}dz+ \int_{\Omega}\frac{|v|^{2^*} \eta^2v_{L}^{2(s-1)}}{d(z, a_{i})^{2^*\beta } }dz\right). \end{aligned} \end{equation} $

(2.14)

对于上式第一项利用不等式$ t^{q}\leq t^{2}+t^{2^*} $ $ (q\in [2, 2^*)) $可得

$ \begin{equation} \frac{|v|^{q}}{d(z, a_{i})^{q\beta }}\leq \frac{v^{2}}{d(z, a_{i})^{2\beta }}+ \frac{|v|^{2^*}}{d(z, a_{i})^{2^*\beta }}. \end{equation} $

(2.15)

从而, 存在$ C_{1} $, $ C_{2}>0 $使得

$ \begin{equation} \begin{aligned} \int _{\Omega }\frac{f(z, d(z, a_{i})^{-\beta } v) \phi}{d(z, a_{i})^{\beta }} dz \leq C_{1}\int _{\Omega }\frac{v^{2}\eta^2 v_{L}^{2(s-1)}}{d(z, a_{i})^{2\beta }}dz+ C_{2}\int_{\Omega}\frac{|v|^{2^*} \eta^2v_{L}^{2(s-1)}}{d(z, a_{i})^{2^*\beta } }dz. \end{aligned} \end{equation} $

(2.16)

最后, 将$ \phi $的表达式代入(2.10)式右边第二项中, 利用$ \mu_{i} $的定义, 中值定理, Young不等式和$ \psi_{i}<1 $可得

$ \begin{equation} \begin{aligned} &\Bigg|\int_{\Omega} \frac{v\phi}{d(z, a_{i})^{2\beta}} \Big(\frac{\mu \prod_{i = 1}^{k}\psi_{i}^2}{\prod_{i = 1}^{k}d(z, a_{i})^{2}}-\frac{\mu_{i}\psi_{i}^{2}}{d(z, a_{i})^{2}} \Big)dz\Bigg| = \Bigg|\int_{\Omega} \frac{\psi_{i}^{2}v^{2}\eta^{2}v_{L}^{2(s-1)}}{d(z, a_{i})^{2(1+\beta)}} \Big( \frac{\mu \prod\limits_{j\neq i, \, j = 1}^{k}\psi_{j}^{2}}{\prod\limits_{j\neq i, \, j = 1}^{k}d(z, a_{j})^{2}}-\mu_{i}\Big)dz\Bigg|\\ \leq& \int_{B_{d}(a_{i}, \rho)} \frac{\psi_{i}^{2}v^{2}\eta^{2}v_{L}^{2(s-1)}}{d(z, a_{i})^{2(1+\beta)}} \Bigg|\frac{\mu }{\prod\limits_{j\neq i, \, j = 1}^{k}d(z, a_{j})^{2}}-\frac{\mu }{\prod\limits_{j\neq i, \, j = 1}^{k}d(a_{i}, a_{j})^{2}}\Bigg|dz\\ \leq& C_{3}\int_{B_{d}(a_{i}, \rho)} \frac{\psi_{i}^{2}v^{2}\eta^{2}v_{L}^{2(s-1)}}{d(z, a_{i})^{2(1+\beta)}}d(z, a_{i})dz = C_{3}\int_{B_{d}(a_{i}, \rho)} \frac{\psi_{i} v \eta v_{L}^{s-1}}{d(z, a_{i})^{1+\beta}} \frac{\psi_{i} v\eta v_{L}^{s-1}}{d(z, a_{i})^{\beta}}dz\\ \leq& \frac{\varepsilon C_{3}}{2}\int_{B_{d}(a_{i}, \rho)} \frac{\psi_{i}^{2}v^{2}\eta^{2}v_{L}^{2(s-1)}}{d(z, a_{i})^{2(1+\beta)}}dz +\frac{C_{3} C_{2}(\varepsilon)}{2}\int_{B_{d}(a_{i}, \rho)} \frac{\psi_{i}^{2}v^{2}\eta^{2}v_{L}^{2(s-1)}}{d(z, a_{i})^{2\beta}}dz, \end{aligned} \end{equation} $

(2.17)

其中$ C_{3} $, $ C_{2}(\varepsilon) $是正常数.

下面利用如下形式的加权Sobolev不等式[3]对(3.17)的右端项做进一步的处理:

$ \begin{equation} \left( \int _{\mathbb {R}^N} \psi^{\tau} \frac{|d(z)^{a} u| ^{2^*(\tau)}}{d(z)^{\tau}} dz \right) ^{\frac{2}{2^*(s)}}\, \leq C(\tau, a, Q)\int _{\mathbb{R}^N} | d(z)^a \nabla _\alpha u|^2 dz, \quad \forall u\in C_0^\infty (\mathbb {R}^N) \end{equation} $

(2.18)

其中$ 0\leq \tau\leq 2 $, $ a> \frac{2-Q}{2} $, $ C(\tau, a, Q)>0 $是常数.从而, 在(2.18)式中取$ \tau = 2 $, $ a = -\beta $, 存在正常数$ C_{4} $使得

$ \begin{equation} \begin{aligned} \int_{B_{d}(a_{i}, \rho)} \frac{\psi_{i}^{2}v^{2}\eta^{2}v_{L}^{2(s-1)}}{d(z, a_{i})^{2(1+\beta)}}dz &\leq C_{4}\int_{B_{d}(a_{i}, \rho)} \frac{|\nabla_{\alpha}(\eta vv_{L}^{s-1})|^{2}}{d(z, a_{i})^{2\beta}}dz\\ &\leq C_{4}\Big(2\int_{B_{d}(a_{i}, \rho)} \frac{v^{2}v_{L}^{2(s-1)}|\nabla_{\alpha}\eta|^{2}}{d(z, a_{i})^{2\beta}}dz+ 2\int_{B_{d}(a_{i}, \rho)} \frac{\eta^{2}v_{L}^{2(s-1)}|\nabla_{\alpha}v|^{2}}{d(z, a_{i})^{2\beta}}dz\\ &\quad+ 2(s-1)^{2}\int_{B_{d}(a_{i}, \rho)} \frac{\eta^{2}v^{2}v_{L}^{2(s-2)}|\nabla_{\alpha}v_{L}|^{2}}{d(z, a_{i})^{2\beta}}dz \Big). \end{aligned} \end{equation} $

(2.19)

因此, 结合(2.17), (2.19)式, (2.10)式右端第二项满足

$ \begin{equation} \begin{aligned} &\Bigg|\int_{\Omega} \frac{v\phi}{d(z, a_{i})^{2\beta}} \Big( \frac{\mu \prod_{i = 1}^{k}\psi_{i}^{2}}{\prod_{i = 1}^{k}d(z, a_{i})^{2}}-\frac{\mu_{i}\psi_{i}^{2}}{d(z, a_{i})^{2}}\Big)dz\Bigg|\\ \leq& (\varepsilon C_{5}+C_{3}(\varepsilon))\int_{B_{d}(a_{i}, \rho)} \frac{v^{2}v_{L}^{2(s-1)}(\eta^{2}+|\nabla_{\alpha}\eta|^{2})}{d(z, a_{i})^{2\beta}}dz\\ &+\varepsilon C_{5}\int_{B_{d}(a_{i}, \rho)} \frac{\eta^{2}v_{L}^{2(s-1)}|\nabla_{\alpha}v|^{2}}{d(z, a_{i})^{2\beta}}dz + \varepsilon(s-1)^{2} C_{5}\int_{B_{d}(a_{i}, \rho)} \frac{\eta^{2}v^{2}v_{L}^{2(s-2)}|\nabla_{\alpha}v_{L}|^{2}}{d(z, a_{i})^{2\beta}}dz. \end{aligned} \end{equation} $

(2.20)

其中$ C_{5} = C_{3}C_{4} $, $ C_{3}(\varepsilon) = \frac{1}{2}C_{3}C_{2}(\varepsilon) $.

将(2.13), (2.16), (2.20)式代入(2.10), 可得

$ \begin{equation} \begin{aligned} &(1-\varepsilon)\int _{\Omega}\frac{\eta ^2v_{L}^{2( s-1) }| \nabla _{\alpha} v| ^2}{d(z, a_{i})^{2\beta}} dz +2( s-1) \int _{\Omega}\frac{\eta^2 v_{L}^{2( s-1) } | \nabla _{\alpha} v_{L}| ^2}{d(z, a_{i})^{2\beta}} dz -C_{1}(\varepsilon)\int _{\Omega} \frac{ v^{2} v_{L}^{2( s-1) } |\nabla_{\alpha}\eta|^{2}}{d(z, a_{i})^{2\beta}} dz\\ \leq& C_{1}\int _{\Omega }\frac{v^{2}\eta^2 v_{L}^{2(s-1)}}{d(z, a_{i})^{2\beta }}dz+ C_{2}\int_{\Omega}\frac{|v|^{2^*} \eta^2v_{L}^{2(s-1)}}{d(z, a_{i})^{2^*\beta } }dz +(\varepsilon C_{5}+C_{3}(\varepsilon))\int_{B_{d}(a_{i}, \rho)} \frac{v^{2}v_{L}^{2(s-1)}(\eta^{2}+|\nabla_{\alpha}\eta|^{2})}{d(z, a_{j})^{2\beta}}dz\\ &+\varepsilon C_{5}\int_{B_{d}(a_{i}, \rho)} \frac{\eta^{2}v_{L}^{2(s-1)}|\nabla_{\alpha}v|^{2}}{d(z, a_{i})^{2\beta}}dz+ \varepsilon(s-1)^{2} C_{5}\int_{B_{d}(a_{i}, \rho)} \frac{\eta^{2}v^{2}v_{L}^{2(s-2)}|\nabla_{\alpha}v_{L}|^{2}}{d(z, a_{i})^{2\beta}}dz\\ \leq& C_{2}\int_{\Omega}\frac{|v|^{2^*} \eta^2v_{L}^{2(s-1)}}{d(z, a_{i})^{2^*\beta } }dz +C_{4}(\varepsilon)\int_{B_{d}(a_{i}, \rho)} \frac{v^{2}v_{L}^{2(s-1)}(\eta^{2}+|\nabla_{\alpha}\eta|^{2})}{d(z, a_{i})^{2\beta}}dz\\ &+\varepsilon C_{5}\int_{B_{d}(a_{i}, \rho)} \frac{\eta^{2}v_{L}^{2(s-1)}|\nabla_{\alpha}v|^{2}}{d(z, a_{i})^{2\beta}}dz + \varepsilon(s-1)^{2} C_{5}\int_{B_{d}(a_{i}, \rho)} \frac{\eta^{2}v_{L}^{2(s-1)}|\nabla_{\alpha}v_{L}|^{2}}{d(z, a_{i})^{2\beta}}dz. \end{aligned} \end{equation} $

(2.21)

取$ \varepsilon = \min\{\frac{1}{1+2C_{5}}, \frac{1}{C_{5}(s-1)}\} $, 则存在常数$ C_{6} $, $ C_7>0 $和$ C_5(\varepsilon)>0 $使得

$ \begin{equation} \begin{aligned} &C_{6}\int _{\Omega}\frac{\eta ^2v_{L}^{2( s-1) }| \nabla _{\alpha} v| ^2}{d(z, a_{i})^{2\beta}} dz +C_{7} \int _{\Omega}\frac{\eta^2 v_{L}^{2( s-1) } | \nabla _{\alpha} v_{L}| ^2}{d(z, a_{i})^{2\beta}} dz\\ \leq& C_{2}\int_{\Omega}\frac{|v|^{2^*} \eta^2v_{L}^{2(s-1)}}{d(z, a_{i})^{2^*\beta } }dz +C_{5}(\varepsilon)\int_{B_{d}(a_{i}, \rho)} \frac{v^{2}v_{L}^{2(s-1)}(\eta^{2}+|\nabla_{\alpha}\eta|^{2})}{d(z, a_{i})^{2\beta}}dz. \end{aligned} \end{equation} $

(2.22)

令$ w = \eta v v_{L}^{s-1} $.将$ w $代入(2.18)式, 利用(2.22)式及$ s>1 $可得

$ \begin{equation} \begin{aligned} \left( \int _{\Omega }\frac{|\eta v v_{L}^{s-1}|^{2^*}}{d(z, a_{i})^{2^*\beta }}dz\right) ^{\frac{2}{2^*}} &\leq C_{8} \int _{\Omega }\frac{| \nabla _\alpha ( \eta v v_t^{s-1})| ^2}{d(z, a_{i})^{2\beta }}dz\\ & \leq 2C_{8}\Bigg( \int _{\Omega } \frac{| \nabla _\alpha \eta |^2 v^2 v_{L}^{2( s-1) }}{d(z, a_{i})^{2\beta }} +\int _{\Omega }\frac{\eta ^2v_{L}^{2( s-1) }| \nabla _\alpha v| ^2}{d(z, a_{i})^{2\beta }}\\ &\qquad +\int _{\Omega }\frac{( s-1) ^2\eta ^2 v_{L}^{2( s-1) } | \nabla _\alpha v_{L}| ^2}{d(z, a_{i})^{2\beta }}dz\Bigg) \\ &\leq sC_{9} \int _{\Omega }\frac{( \eta ^2+| \nabla _\alpha \eta | ^2)v^2v_{L}^{2(s-1)}} {d(z, a_{i})^{2\beta }}z + C_{10}s \int _{\Omega }\frac{\eta ^2| v| ^{2^*}v_{L}^{2(s-1)}}{d(z, a_{i})^{2^*\beta }}dz. \end{aligned} \end{equation} $

(2.23)

选取$ q\in (\frac{Q}{2}\, , \, \frac{Q(Q-2)}{2(Q-2-2\sqrt{(\frac{Q-2}{2})^{2}- \sum\limits^{k}_{i = 1}\mu_{i} })} ) $, 使其满足$ (2^*-2)q<\frac{2Q}{Q-2-2\sqrt{(\frac{Q-2}{2})^{2}- \sum\limits^{k}_{i = 1}\mu_{i}}}, 2<\frac{2q}{q-1}<2^*. $类似文[8]的计算方法, 对任意的$ \varepsilon >0 $, 存在常数$ C_{11}>0 $使得

$ \begin{equation} \int _{\Omega }\frac{\eta ^2| v| ^{2^*}v_{L}^{2(s-1)}}{d(z, a_{i})^{2^*\beta }}dz \leq C_{11}\varepsilon^2 \left( \int _{\Omega }\Big|\frac{ \eta v v_{L}^{s-1}}{d(z, a_{i})^{\beta }}\Big|^{2^*}dz \right) ^{\frac{2}{2^*}}+ C_{11}\varepsilon^{-\frac{2Q}{2q-Q}}\int _{\Omega }\Big|\frac{\eta v v_{L}^{s-1}}{d(z, a_{i})^{\beta }}\Big|^2 dz. \end{equation} $

(2.24)

将(2.24)式代入(2.23)式中并取$ \varepsilon = \frac{1}{\sqrt{2C_{10}C_{11}s}} $, 则有

$ \begin{equation} \left( \int _{\Omega }\frac{| \eta v v_{L}^{s-1}|^{2^*}}{d(z, a_{i})^{2^*\beta }}dz \right) ^{\frac{2}{2^*}} \leq C_{12}s^{\frac{2q}{2q-Q}}\int _{\Omega } \frac{( \eta^2+| \nabla _\alpha \eta | ^2)| v|^2v_{L}^{2(s-1)}}{d(z, a_{i})^{2\beta }}dz, \end{equation} $

(2.25)

其中$ C_{12} = C_{9}+(C_{10}C_{11})^{\frac{2q}{2q-Q}}>0 $.在(2.25)式中应用$ v_{L} $和$ \eta $的性质可得

$ \begin{equation} \begin{aligned} \left( \int _{B_{d}(a_{i}, r)} \frac{|v|^{2}v_{L}^{2^*s-2}}{d(z, a_{i})^{2^*\beta }}dz\right)^{\frac{2}{2^*}} &\leq \left( \int _{\Omega }\frac{\eta ^{2^*}| v|^{2}v_{L}^{2^*s-2}}{d(z, a_{i})^{2^*\beta }}\, dz\right) ^{\frac{2}{2^*}}\\ &\leq C_{12}s^{\frac{2q}{2q-Q}}\int _{\Omega } \frac{( \eta^2+| \nabla _\alpha \eta |^2)| v|^2v_{L}^{2(s-1)}}{d(z, a_{i})^{2\beta }}dz\\ &\leq \frac{C_{12}r^{2^*-2}s^{\frac{2q}{2q-Q}}}{( \rho -r) ^2} \int _{B_{d }(a_{i}, \rho)}\frac{( \eta^2+| \nabla _\alpha \eta| ^2)|v| ^2v_{L}^{2(s-1)}}{d(z, a_{i})^{2^*\beta }}dz\\ &\leq C_{13}s^{\frac{2q}{2q-Q}} \int _{B_{d }(a_{i}, \rho)} \frac{|v|^{2}v_{L}^{2s-2}}{d(z, a_{i})^{2^*\beta }}dz. \end{aligned} \end{equation} $

(2.26)

取$ s_{0} \in\Big(\frac{Q}{Q-2}, \frac{Q}{Q-2-2\sqrt{(\frac{Q-2}{2})^{2}-\sum\limits_{i = 1}^{k}\mu_{i}}}\Big) $.则计算可得$ \frac{2Q}{Q-2}<2s_{0}< \frac{2Q}{Q-2-2\sqrt{(\frac{Q-2}{2})^{2}-\sum\limits_{i = 1}^{k}\mu_{i}}} $.从而, 由引理2.2知, 对任意的$ u\in D^{1, 2}_{\alpha}(\Omega) $有$ \int_{\Omega}|u|^{2s_{0}}dz<+\infty $.

取序列$ s_j = s_{0}( \frac{2^*}{2})^j>1 $, $ \forall j = 0, 1, 2, \cdots $, 将$ s_{j} $代入(2.26)式, 利用Moser迭代技巧得

$ \begin{equation} \begin{aligned} &( \int _{B_{d}(a_{i}, r)} \frac{|v|^{2}v_{L}^{2s_{j+1}-2}}{d(z, a_{i})^{2^*\beta }}dz ) ^{\frac{1}{2s_{j+1}}}\\ \leq& C_{13}^{\sum\limits_{k = 0}^j \frac{1}{2s_k}} (s_{0})^{\frac{q}{s_{0}(2q-Q)}\sum\limits_{k = 0}^{j} (\frac{2}{2^*})^{j}} (\frac{2^*}{2})^{\frac{q}{s_{0}(2q-Q)}\sum\limits_{k = 0}^{j} j(\frac{2}{2^*})^{j}} \times( \int _{B_{d}(a_{i}, \rho)} \frac{|v|^{2}v_{L}^{2s_{0}-2}}{d(z, a_{i})^{2^*\beta }}dz )^{\frac{1}{2s_{0}}}. \end{aligned} \end{equation} $

(2.27)

因为级数$ \sum\limits_{k = 0}^{\infty} \frac{1}{2s_k} $, $ \sum\limits_{k = 0}^{\infty} (\frac{2}{2^*})^{j} $, $ \sum\limits_{k = 0}^{\infty} j(\frac{2}{2^*})^{j} $是收敛的, 所以, 对(2.27)式取极限$ j\to\infty $, 可知:存在不依赖于$ L $的常数$ C>0 $使得$ \| v\| _{L^{\infty }(B_{\rho _0})}\leq C $.因此, 由$ v $的定义可得不等式(1.12)成立, 进而可得(1.13)式亦成立.定理1.1得证.